# ME statics shear/moment diagrams through integration

1. Nov 7, 2014

### Feodalherren

1. The problem statement, all variables and given/known data

2. Relevant equations
w=(2/3)x

3. The attempt at a solution
So integrating w to get V

(1/3)x^2 +C

Then Sum in the Y-direction should be
9-(2/3)x+18=0

Somebody tell me what in god's name is going on here. It seems like they are ignoring the 18kN force and then plugging in my shear in the sum of Y-forces... What?!

This totally contradicts the formula dV/dx = -w

That would simply give me V=(1/3)x^2 + C

Last edited: Nov 7, 2014
2. Nov 7, 2014

### SteamKing

Staff Emeritus
It would help if you would post the complete problem statement, as the rules of PF ask you to do.

3. Nov 8, 2014

### SteamKing

Staff Emeritus
The equation for ΣFy = 9 - (1/3)x2 - V = 0, applies to the free body diagram depicted in Fig. (b). V and M are the shear force and bending moment which keep this FB in equilibrium under the applied load and the reaction at the LH end of the beam. The right hand reaction of 18 kN is not involved in maintaining this equilibrium, except when dealing with the loaded beam as a whole.

This is an incorrect statement. The loading of the beam, w = (2/3) x kN/m, is clearly expressing the magnitude of the distributed loading as a function of position along the beam, not the total applied load between 0 and x. Your statement is mixing the two reactions, which are both measured in kN, with a distributed loading, which is measured in kN/m, and thus has no meaning.

4. Nov 9, 2014

### Feodalherren

But what about the fact that dV/dx = -w

That would mean that V= (1/3)x^2
??

5. Nov 9, 2014

### SteamKing

Staff Emeritus
According to the OP, V = 9 - (1/3)x^2. What's dV/dx in this case? How does it compare to w = (2/3)x as shown in Fig. (b) of the OP?